1. ## [SOLVED] Limits

this part of finding the limit [in this form] ALWAYS confuses me, i really can get it [maybe the (h) throws me off]

the limit of [f(x+h)-f(x)] / h
1. f(x) = 3x-1

How does 3x-1 substitute in when there is an "h", it is confusing

2. Hello,
Originally Posted by >_<SHY_GUY>_<
this part of finding the limit [in this form] ALWAYS confuses me, i really can get it [maybe the (h) throws me off]

the limit of [f(x+h)-f(x)] / h
1. f(x) = 3x-1

How does 3x-1 substitute in when there is an "h", it is confusing
$f({\color{red}x})=3{\color{red}x}-1$

$f({\color{red}x+h})=3({\color{red}x+h})-1$

So $\frac{f(x+h)-f(x)}{h}=\frac{[3(x+h)-1]-[3x-1]}{h}=\frac{3x+3h-1-3x+1}{h}=\frac{3h}{h}$

Simplify

3. $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{[3(x+h)-1]-(3x-1)}{h}=\lim_{h\to 0}\frac{3x+3h-1-3x+1}{h}=$

$=\lim_{h\to 0}\frac{3h}{h}=3$

4. Originally Posted by Moo
Hello,

$f({\color{red}x})=3{\color{red}x}-1$

$f({\color{red}x+h})=3({\color{red}x+h})-1$

So $\frac{f(x+h)-f(x)}{h}=\frac{[3(x+h)-1]-[3x-1]}{h}=\frac{3x+3h-1-3x+1}{h}=\frac{3h}{h}$

Simplify
hmm...right...but it says [and i forgot to mention that x approaches zero [the limit] the answer 3?

5. nevermind....errrrr im so dumb...thank you, i get it now haha